J.R. S. answered 11/27/17
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Ph.D. in Biochemistry--University Professor--Chemistry Tutor
Assuming the 0.9% NaCl is mass/volume (which it usually is, as in normal saline), then you have 0.9 g NaCl/100 ml.
Since 3 L is equivalent to 3000 ml, you have the following:
0.9 g/100 ml x 3000 ml = 27 g NaCl in 3 Liters
Converting this to mg, you have 27 g x 1000 mg/g = 27,000 mg NaCl in 3 liters of the solution.